Solve (80/3-25)y+(28-25)y_{6}=280

$$( \frac { 80 } { 3 } - 25 ) y + ( 28 - 25 ) y _ { 6 } = 280$$

Solve for y

$$y=-\frac{9y_{6}}{5}+168$$

Steps for Solving Linear Equation

Subtract 25 from \frac{80}{3} to get \frac{5}{3}.

$$\frac{5}{3}y+\left(28-25\right)y_{6}=280$$

Subtract 25 from 28 to get 3.

$$\frac{5}{3}y+3y_{6}=280$$

Subtract 3y_{6} from both sides.

$$\frac{5}{3}y=280-3y_{6}$$

Divide both sides of the equation by \frac{5}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{3}y}{\frac{5}{3}}=\frac{280-3y_{6}}{\frac{5}{3}}$$

Dividing by \frac{5}{3} undoes the multiplication by \frac{5}{3}.

$$y=\frac{280-3y_{6}}{\frac{5}{3}}$$

Divide 280-3y_{6} by \frac{5}{3} by multiplying 280-3y_{6} by the reciprocal of \frac{5}{3}.

$$y=-\frac{9y_{6}}{5}+168$$

Solve for y_6

$$y_{6}=-\frac{5y}{9}+\frac{280}{3}$$

Steps for Solving Linear Equation

Subtract 25 from \frac{80}{3} to get \frac{5}{3}.

$$\frac{5}{3}y+\left(28-25\right)y_{6}=280$$

Subtract 25 from 28 to get 3.

$$\frac{5}{3}y+3y_{6}=280$$

Subtract \frac{5}{3}y from both sides.

$$3y_{6}=280-\frac{5}{3}y$$

The equation is in standard form.

$$3y_{6}=-\frac{5y}{3}+280$$

Divide both sides by 3.

$$\frac{3y_{6}}{3}=\frac{-\frac{5y}{3}+280}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y_{6}=\frac{-\frac{5y}{3}+280}{3}$$

Divide 280-\frac{5y}{3} by 3.

$$y_{6}=-\frac{5y}{9}+\frac{280}{3}$$